hello
I couldn't find previous posts to answer this, but please point me to any.
I am trying to understand bsts, starting with no regressors
Here is code which appears to mimic bsts, producing graphs similar to the
model plot, but gives a rather different posterior distribution for the
parameter
p172, another reason why I was puzzled that the default
estimate used dev instead.
thanks
Greg
> On 05/02/2014 12:56, Greg Dropkin wrote:
>> thanks Simon
>> also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minim
h
> these sort of data, and looking back at my list of open issues I see
> that `someone' was you as well! It's possible that moving to a Pearson
> estimator of the scale will solve that problem too.
>
> Thanks for this... very helpful.
>
> best,
> Simon
>
>
roximation breaks down. So yes, the
> Pearson estimator is probably a better bet in the latter case.
>
> best,
> Simom
>
> ps. Note that the chi squared approximation for the *difference* in
> deviance between two nested models does not suffer from this problem.
>
>
> O
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd a
just to clarify how I see the error, it was the mis-definition of the
penalty term in the function dv. The following code corrects this error.
What is actually being minimised at this step is the penalised deviance
conditional on the smoothing parameter. A second issue is that the optim
default ("N
hi
probably a silly mistake, but I expected gam to minimise the penalised
deviance.
thanks
greg
set.seed(1)
library(mgcv)
x<-runif(100)
lp<-exp(-2*x)*sin(8*x)
y<-rpois(100,exp(lp))
plot(x,y)
m1<-gam(y~s(x),poisson)
points(x,exp(lp),pch=16,col="green3")
points(x,fitted(m1),pch=16,cex=0.5,col="bl
please ignore this, I see the error.
greg
> hi
>
> probably a silly mistake, but I expected gam to minimise the penalised
> deviance.
>
> thanks
>
> greg
>
> set.seed(1)
> library(mgcv)
> x<-runif(100)
> lp<-exp(-2*x)*sin(8*x)
> y<-rpois(100,exp(lp))
> plot(x,y)
> m1<-gam(y~s(x),poisson)
> points
hi
I'm trying to understand (a little) the code behind summary.gam, and have
the Biometrika article referred to in the Help. But am stuck early on.
In the code, starting at line 167:
if (est.disp)
rdf <- residual.df
else rdf <- -1
res <- testStat(p, Xt, V, df[i], type = p.type,
res.df = rdf)
=TRUE)$val
> ldsq<-sum(log(eigSq[1:16]))
> F3q<-(ldhsq-ldsq)/2
>
> #4th term
> Mp=3
> F4q<-Mp/2*log(2*pi*phiq)
>
> F1q+F3q-F4q
> m3$gcv
>
> #quite different
>
> #but if phiq is replaced by the Pearson est
sum(m3$edf))
F1p<-Dpq/(2*phip)
F3p<-F3q
#third term independent of scale
F4p<-Mp/2*log(2*pi*phip)
F1p+F3p-F4p
m3$gcv
#closer but still wrong
#is there a value of phi which makes this work?
f<-function(t) (Dpq/(2*t) + F3q + Mp/2*log(2*pi*t) - m3$gcv)^2
optimise(f,interval=c(0.5,1.5))$min-&
hi Simon
this follows on from the example where gcv increased unexpectedly with
increasing basis dimension. I'm now looking at t2 tensor splines with
REML, and find that the REML score can increase when adding a new
predictor. Again, this seems odd.
thanks
Greg
library(mgcv)
set.seed(0)
#simula
thanks Simon
I'll upgrade R to try t2. The data I'm actually analysing requires scaled
Poisson so I don't think REML is an option.
thanks
Greg
On 14/02/12 11:22 Simon Wood wrote:
That's interesting. Playing with the example, it doesn't seem to be a
local minimum. I think that this happens beca
hi
Using a ts or tprs basis, I expected gcv to decrease when increasing the
basis dimension, as I thought this would minimise gcv over a larger
subspace. But gcv increased. Here's an example. thanks for any comments.
greg
#simulate some data
set.seed(0)
x1<-runif(500)
x2<-rnorm(500)
x3<-rpois(50
re: Cubic splines in package "mgcv"
I don't have access to Gu (2002) but clearly the function R(x,z) defined
on p126 of Simon Wood's book is piecewise quartic, not piecewise cubic.
Like Kunio Takezawa (below) I was puzzled by the word "cubic" on p126.
As Simon Wood writes, this basis is not actu
hadn't realised the answer would be in the source code!
anyway, this appears to work. The only difference is in the last section.
greg
--
library(mgcv)
#simulate some data
x1<-runif(500)
x2<-rnorm(500)
x3<-rpois(500,3)
d<-runif(500)
t<-runif(500,20,50)
linp<--6.5+x1+2*x2-x3+2*exp(-2*d)*sin(2*p
hello
I'm learning mgcv and would like to obtain numerical output corresponding
to plot.gam.
I can do so when seWithMean=FALSE (the default)
but only approximately when seWithMean=TRUE.
Can anyone show how to obtain the exact values?
Alternatively, can you clarify the explanation in the manual
re [R] matrix^(-1/2)
re the discussion in November on this thread. I don't know about expm but
the problem must be equivalent to solve(B^(1/2)) and a solution will exist
iff B is invertible and has a square root A with A%*%A = B. For 2x2
matrices necessary and sufficient conditions for B to have a
] [ author ]
On Wed, 23 Sep 2009, Greg Dropkin wrote:
> hi
>
> sorry if this has been discussed before, but I'm wondering why the scaled
> Schoenfeld residuals do not follow the defining formula for obtaining them
> from the ordinary Schoenfeld residuals, but are instead offse
reason for the offset
#or am I missing something?
thanks
Greg
Greg Dropkin
gr...@gn.apc.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.ht
re post from
bkelcey at umich.edu bkelcey at umich.edu
Wed Feb 27 15:09:48 CET 2008
If the response y is given as the proportion of successes out of n trials,
and y, n, p, x, and z are vectors of length M, and the model is logit(p) =
b0 + b1*x + b2*z then for the score test for the null hypothesi
21 matches
Mail list logo
